Search any question & find its solution
Question:
Answered & Verified by Expert
Chlorine is prepared in the laboratory by treating manganese dioxide \(\left(\mathrm{MnO}_2\right)\) with aqueous hydrochloric acid according to the reaction \(4 \mathrm{HCl}(\mathrm{aq})+\mathrm{MnO}_2(\mathrm{~s}) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(\ell)+\mathrm{MnCl}_2(\mathrm{aq})+\mathrm{Cl}_2(\mathrm{~g})\) How many grams of \(\mathrm{HCl}\) react with \(5.0 \mathrm{~g}\) of manganese dioxide?
Solution:
2054 Upvotes
Verified Answer
\(1 \mathrm{~mol}\) of \(\mathrm{MnO}_2\), i.e., \(55+32=87 \mathrm{~g}\) of \(\mathrm{MnO}_2\) react with 4 moles of \(\mathrm{HCl}\), i.e., \(4 \times 36.5 \mathrm{~g}=146 \mathrm{~g}\) of \(\mathrm{HCl}\).
Hence \(5.0 \mathrm{~g}\) of \(\mathrm{MnO}_2\) will react with \(\mathrm{HCl}\)
\(=\frac{146}{87} \times 5.0 \mathrm{~g}=8.40 \mathrm{~g}\)
Hence \(5.0 \mathrm{~g}\) of \(\mathrm{MnO}_2\) will react with \(\mathrm{HCl}\)
\(=\frac{146}{87} \times 5.0 \mathrm{~g}=8.40 \mathrm{~g}\)
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.